If the domain of the function $f(x)=\frac{\sqrt{x^{2}-25}}{(4-x^{2})} +\log _{10}(\mathrm{x}^{2}+2 \mathrm{x}-15)$ is $(-\infty, \alpha) \cup[\beta, \infty)$, then $\alpha^{2}+\beta^{3}$ is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Sets Relations and Functions

If the domain of the function

f (x) = \frac{x ^{2} - 25}{( 4 - x ^{2} )} + lo g_{10} (x^{2} + 2 x - 15)

(- \infty, α) \cup [β, \infty)

, then

α^{2} + β^{3}

is equal to :

Answer: (C)

f (x) = \frac{x ^{2} - 25}{4 - x ^{2}} + lo g_{10} (x^{2} + 2 x - 15)

.

Square-root condition:

x^{2} - 25 \geq 0 ⟹ x \leq - 5

x \geq 5

.

Denominator condition:

4 - x^{2} \neq = 0 ⟹ x \neq = \pm 2

(already outside the region above).

Log condition:

x^{2} + 2 x - 15 > 0 ⟹ (x + 5) (x - 3) > 0 ⟹ x < - 5

x > 3

.

Intersecting

{x \leq - 5 or x \geq 5}

with

{x < - 5 or x > 3}

For $x \leq - 5$ : combined with $x < - 5$ (since $x = - 5$ fails the strict log inequality) gives $x < - 5$ .
For $x \geq 5$ : this already satisfies $x > 3$ , so all of $x \geq 5$ survives.

Domain

= (- \infty, - 5) \cup [5, \infty)

, so

α = - 5

β = 5

α^{2} + β^{3} = 25 + 125 = 150.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option